Codeforce940C (Phone Numbers)

C. Phone Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that p_i < q_i and for all j < i it is satisfied that p_j = q_j. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

Copy

3 3
abc

Output

aca

Input

Copy

3 2
abc

Output

ac

Input

Copy

3 3
ayy

Output

yaa

Input

Copy

2 3
ba

Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac,

aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.

分析：题意是找出按字典序的下一个长度为k的字符串。

当n<k时，在原字符串加上字典序最小的字符即可；

当n>=k时，贪心，从k-1到0找到不是最大字符的一位j，然后s[j]赋值为按字典序的下一个字符，

j以后的全部为赋值为字典序最小的字符。

#include<cstdio>
#include<cstring>
using namespace std;
char s[200000];
int b[30];
char a[30];
int main()
{
    int N,k;
    scanf("%d%d",&N,&k);
    scanf("%s",s);
    for(int i=0;i<N;i++)
    b[s[i]-'a']++;
    
    int cnt=0;
    for(int i=0;i<26;i++)
    if(b[i]) a[cnt++]='a'+i;
    cnt--;
    if(k>N)
    {
        s[k]='';
        for(int i=N;i<k;i++) s[i]=a[0];
        printf("%s
",s);
    }
    else//找出那个位置 
    {
        s[k]='';
        for(int i=k-1;i>=0;i--)
        {
            if(s[i]==a[cnt]) {s[i]=a[0];continue;}
            else
            {
                int j=0;
                for(;j<=cnt;j++)
                {
                    if(s[i]<a[j]) break;
                }
                s[i]=a[j];
                break;
            }
        }
        printf("%s
",s);
    }
    return 0;
}